关于连续函数空间中交集集合基数的求解问询

Hey there, let's work through this problem step by step to figure out the cardinality of the set in question.

First, let's recap the setup clearly:

• We have the linear functional $F: (C[0,1], |\cdot|_\infty) \to \mathbb{R}$ defined by $F(f) = \int_0^1 f(t)dt$.
• $X$ is the subspace of continuous functions on $[0,1]$ with $f(0)=f(1)=0$.
• $S_X$ is the unit sphere in $X$, i.e., ${f \in X : |f|_\infty =1}$.

We want to find the size of $S_X \cap {f \in X : F(f) = |F|}$.

Step 1: Confirm $|F| =1$ #

First, we know for any $f \in X$, $|F(f)| = \left|\int_0^1 f(t)dt\right| ≤ \int_0^1 |f(t)|dt ≤ |f|_\infty \cdot 1$, so $|F| ≤1$.

To see $|F|=1$, consider a sequence of functions $f_n \in X$ where:

• $f_n(t) = nt$ for $0 ≤ t ≤ 1/n$,
• $f_n(t)=1$ for $1/n ≤ t ≤ 1-1/n$,
• $f_n(t)=n(1-t)$ for $1-1/n ≤t ≤1$.

Each $f_n$ has $|f_n|\infty=1$, and $F(f_n) = 1 - \frac{1}{n}$. As $n\to\infty$, $F(f_n)\to1$, so the supremum of $|F(f)|/|f|\infty$ over non-zero $f\in X$ is exactly 1. Thus $|F|=1$.

Step 2: Prove no such function exists in the intersection #

Suppose, for contradiction, there exists an $f \in S_X \cap {f \in X : F(f)=|F|}$. That means:

1. $|f|_\infty=1$, so $f(t) ≤1$ for all $t\in[0,1]$,
2. $F(f)=1$, so $\int_0^1 f(t)dt=1$,
3. $f(0)=f(1)=0$.

Consider the function $g(t) = 1 - f(t)$. This function is continuous (since $f$ is continuous), non-negative (because $f(t)≤1$ everywhere), and:
$$\int_0^1 g(t)dt = \int_0^1 1 dt - \int_0^1 f(t)dt =1-1=0.$$

For non-negative continuous functions, the integral over $[0,1]$ is zero if and only if the function is identically zero. So $g(t)=0$ for all $t\in[0,1]$, which implies $f(t)=1$ for all $t$. But this contradicts $f(0)=0$ (since $f$ must be in $X$).

Conclusion #

There are no functions that satisfy all the required conditions. Therefore, the set $S_X \cap {f \in X : F(f)=|F|}$ is empty, so its cardinality is 0.

备注：内容来源于stack exchange，提问作者Buton Hao